∫\(\frac {5(x^6+1)}{X+1}\)dx = (where C is a constant of integration.)
∫\(\frac {5(x^6+1)}{X+1}\)dx = (where C is a constant of integration.)
\(\frac {5x^7}{7}\)+ 5x + 5 tan-1 x + c
5 tan–1 x + log (x2 + 1) + C
5(x + 1) + log (x + 1) + C
x5 – \(\frac {5x^3}{3}\) + 5x + C
The Correct Option is D
Solution and Explanation
Let I = ∫\(\frac {5(x^6+1)}{X+1}\)dx
I = 5 ∫ \(\frac {(x^2)^3+(1)^3}{X+1}\)dx
I =5 ∫ \(\frac {(x^2 +1) (x^4 -x^2 +1)}{(X^2+1)}\)dx
I =5 ∫ (x4 -x2 +1) dx
I =5 ∫ x4 dx - 5 ∫ x2 dx + 5 ∫ 1 dx
I = 5 x \(\frac {x^5}{5}\) - 5\(\frac {x^3}{3}\) + 5x + C
I = x5 – \(\frac {5x^3}{3}\) + 5x + C
Therefore, the correct option is (D) x5 – \(\frac {5x^3}{3}\) + 5x + C
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Concepts Used:
Integration by Parts
Integration by Parts is a mode of integrating 2 functions, when they multiplied with each other. For two functions ‘u’ and ‘v’, the formula is as follows:
∫u v dx = u∫v dx −∫u' (∫v dx) dx
- u is the first function u(x)
- v is the second function v(x)
- u' is the derivative of the function u(x)
The first function ‘u’ is used in the following order (ILATE):
- 'I' : Inverse Trigonometric Functions
- ‘L’ : Logarithmic Functions
- ‘A’ : Algebraic Functions
- ‘T’ : Trigonometric Functions
- ‘E’ : Exponential Functions
The rule as a diagram:
